Answer:
Explanation:
The mass of the steel ball, m₁ = 3.0 kg
The mass of the arm, m₂ = 4.0 kg
The length of the arm, r = 7 cm
r = 7/100
r = 0.7m
The angle between the shoulder and the arm = θ
The magnitude of the torque is given by the formula:
[tex]\tau=Fr\sin \theta[/tex]
The force, F, is the combined weight of the steel ball and the arm
F = (m₁ + m₂)g
F = (3.0 + 4.0)(9.8)
F = 7(9.8)
F = 68.6N
a) The magnitude of the gravitational force about the athlete's shoulder if he holds his arm straight parallel to the floor
θ = 180°
[tex]\begin{gathered} \tau\text{ = 68.6(0.7)}\sin 180 \\ \tau=48.02(0) \\ \tau\text{ = 0}Nm \end{gathered}[/tex]
b) 45° below horizontal
θ = 45°
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